Question

составьте формулу функции,если известно,что графиком этой функции являестся прямая,проходящая через начало координат и точку B(12;-3) Плз напишите решение и ответ: y= Ау помогите

Answer 1

График пряммая, значит функция линейная и имеет вид y=kx+b;

 

начало координат (0;0)

Так как искомая пряммая проходит через точки (0;0) и B(12;-3), то

0=kx+b;

-3=12k+b;

 

b=0;

-3=12k+b;

 

b=0;

-3=12k

 

b=0;

k=-3:(12)

 

b=0

k=-0.25

 

т.е. уравнение функции имеет вид y=-0.25x

Answer 2

у=-3

х=12

у=кх

к=-4

у=-0.25х

Answer 3

Function Formula Given Graph as a Line Passing Through the Origin and Point B(12, -3)

To find the formula of the function when the graph is a straight line passing through the origin and the point B(12, -3), we can use the point-slope form of a linear equation. The point-slope form is given by y - y1 = m(x - x1), where m is the slope of the line and (x1, y1) is a point on the line.

Solution and Answer

The equation of the line passing through the origin and point B(12, -3) can be expressed in the point-slope form as:

y - (-3) = (3 - (-3))/(12 - 0) * (x - 12)

Simplifying this equation gives:

y + 3 = (3 + 3)/12 * (x - 12)

y + 3 = 6/12 * (x - 12)

y + 3 = 1/2 * (x - 12)

y + 3 = 1/2x - 6

y = 1/2x - 6 - 3

y = 1/2x - 9

Therefore, the formula of the function is y = 1/2x - 9.

This formula represents the function whose graph is a straight line passing through the origin and the point B(12, -3).

I hope this helps! If you have further questions or need additional assistance, feel free to ask.